Question

You are measuring a line with a steel tape that has a standardized length equal to 100.000 feet and a weight of 2.25 lb. Three full tape lengths and one partial length were used when making the measurement, resulting in a recorded length of 389.28 ft. The tape was supported at ends only and a pull of 12 pounds was used. What is the distance after being corrected for sag?

Answer 1

Ans) We know,

Correction due to sag, Cs = - L / 24

where, W = Weight of tape = 2.25 lb

L = measured length = 389.28 ft

P = Applied pull = 12 lb

Putting values,

=> Cs = - () (389.28) / (24 x )

=> Cs = - 0.5702 ft

Hence, distance after correction = 389.28 - 0.5702 = 388.7098 ft 388.71 ft